[koube]

Forgive me if this is a dumb question, but isn't every vote total extremely unlikely if you make it precise to the exact number of votes? Like the chances of getting n+1, n+2... votes is roughly the same probability.

For example the probability of getting [1,2,3,4,5,6] as the winning numbers in the lottery is the same as any random set of numbers.

[jvanderbot]

The question was "How likely is it that the votes worked out so well that they were basically even 1/10 percentages and not ugly numbers?"

So for a given number of votes, which determines a split, how many times does the split come out so nice? Answer: Effectively none - there are always ugly numbers with lots of decimal places.

Now that analysis comes after they conjecture that the percentages were fixed apriori. The first comment "That seems fishy" basically says this. "How can it be that we're so close to even 1/10 percentages. How can it be that we're exactly one vote off from nice 1/10 percentages"? Fishy indeed - must be rounding.

And they tell you: it's very unlikely to be 1 vote off from nice 0.1% percentage splits.

[tumult]

Another way of writing it out:

How likely is it that you'd get these votes distributions

    51.2000000%
    44.2000000%
    04.6000000%

exactly? With all of those clean 0s? Very low.

But it's also possible that there was sloppy reporting and the vote counts were re-processed at some point in the chain and rounded to one decimal place.

[contravariant]

Well there weren't zeros but within rounding error it was exact.

That actually gives a way to estimate the probability. There's 1002 choose 2 ways to divide 1000 permils over the 3 options. While there's 10 058 776 choose 2 ways to divide the 10 058 774 votes. That works out to about 1e-8 of the possible results being an exact multiple of 0.1% up to rounding error.

Of course an actual election doesn't simply pick one of the possible results at random (heck even if everyone voted randomly that wouldn't be the case). However these 'suspicious' results are distributed in a very uniform stratified fashion, any probability distribution that's much wider than 0.1% would approximately result in the same 1e-8 probability. And pretty much no reasonable person would expect a priori that the vote would result in such a suspicious number with such a high accuracy, so this should be considered strong evidence of fraud to most people.

[neura]

It's more that if you start with those clean, single decimal percentages and a total number of votes, you'd end up with decimals for number of votes, which isn't possible. So if you then remove the decimal from the votes, you get slightly different percentage values when taken to 7 decimal places, but the original decimals would still be the same.

The chances of those numbers occurring normally for all 3 vote counts together is just ridiculously tiny.

[culi]

Actually if you're going that many decimals its

  51.1999971%
  44.1999989%
  04.6000039%

[achempion]

The numbers are definitely sus, but what if they were

    52.2543689%
    44.2689426%
    04.6345625%

How likely is it that you'd get these votes distributions exactly with these exact tails? Compared to all other possibilities?

[kadoban]

Basically the same 1/verymany chance, but that doesn't matter. The difference is that there's no particular reason to choose these numbers to start with. There _is_ a reason to choose nice round, but not too round numbers: that's what humans do.

[jvanderbot]

The question is "How likely is it that humans would begin with those numbers when fudging?" answer is low.

But clean numbers? Much more likely.

[happyopossum]

But that’s not what happened, if you read the article it actually expands everything out to the seventh decimal, and they’re not all zeros

[jvanderbot]

Yes, they are not all zeros, but they are exactly what you'd expect if someone picked percentages that were all zeros, then added +1/-1 to get integer votes.

So the argument is once removed, but still compelling.

[tumult]

It is equivalent to all zeroes with the numeric precision used.

[noitpmeder]

This comment perfectly distills this post.

[melenaboija]

The same probability as any combination of three results with 7 decimals and adding up to 100.

[staunton]

The question isn't "probability you get those exact numbers". It's "probability that you get numbers which all have at most N decimal places".

[enoch_r]

If the lottery administrator's daughter wins the lottery, he may say "no, no - don't you see, her probability of getting the winning numbers is exactly the same as anyone else's!"

But in reality, we can say that:

- her probability of winning in the world where her father is cheating is very high

- her probability of winning in the world where her father isn't cheating is very low

Together these two facts give us evidence about which world we're actually inhabiting - though of course we can never be completely certain!

In the same way, yes, it's equally (im)probable that the winning percent will be 51.211643879% or 51.200000000%. But the latter is more likely to occur in a world where Maduro said "get me 51.2% of the votes" and someone just did that mechanically with a pocket calculator, which is good evidence about which world we live in.

[ChadNauseam]

The other commenters point at the explanation but don't explain it rigorously IMO. Here's how I'd say it.

60% is a nice, round percentage. In an honest election, this is just as likely to be reported as any nearby percentage, like 59.7% or 60.3%. As you mention, any particular percentage is equally (and extremely) unlikely. SUppose this you estimate the chance of this occurring, given an honest election, is 1/1000.

60% however is a much more likely outcome if the election results were faked sloppily. A sloppy fake is reasonably likely to say "Well, why not just say we won 60%". Suppose you estimate the chance of this occurring, given a sloppily faked election, are 1/100.

Bayes' theorem tells us that we can use this information to "update our beliefs" in favor of the election being faked sloppily and away from the election being honest. Say we previously (before seeing this evidence) thought the honest:faked odds were 5:1. That is, we felt it was 5 times more likely that it was honest than that it was sloppily faked. We can then multiply the "honest" by 1/1000 (chance of seeing this if it was honest), and the "faked" by 1/100 (chance of seeing this if it was faked), to get new odds of (5 * 1/1000):(1 * 1/100), which simplifies to 1:2. So in light of the new evidence, and assuming these numbers that I made up, it seems twice as likely that the election was faked.

This exact analysis of course relies on numbers I made up, but the critical thing to see here is that as long as we're more likely to see this result given the election being faked than given it being honest, it is evidence of it being faked.

[a0123]

Yeah, they just forgot to report 59.869280705993% instead of 60%. They would have got away with it too, if it weren't for those cunning statisticians. They just forgot to come up with a random, credible number. Happens to the best of us I guess.

To think they could have got away with it if only they hadn't forgotten.

That´s what you get when you defer the dirty work to interns on their first day, I guess. Which you always rely on to stay in power. Wouldn't want to rely on competent advisers who would have reminded you to come up with a non-round number with 8 or 9 decimals.

[marcosdumay]

> They would have got away with it too

Well, on this case they wouldn't. The smoking gun is their refusal to publish the counting totals, the round ratio is just some extra confirmation.

[paxys]

The second half of the article answers this very question.

Here's an example – if I generate 10 random numbers between 1 and 100, what is more likely: all ten are multiples of 10, or at least one is not a multiple of 10?

[Waterluvian]

Yes. For one set. But if your next lottery is 4,5,6,7,8,9 and then 11,12,13,14,15,16 it becomes improbable.

The issue here is that a bunch of the percentages imply super round numbers.

The signal isn’t that there’s a round number. It’s that they’re all round numbers.

[neura]

I've tried to explain this a couple of times, but I keep falling back on the calculations used to show the problem (that it's not the numbers themselves, but the pattern). This comment nailed it with simply "It's that they're all round numbers". I've always been terrible at rephrasing things to make stronger points in a more concise way. Thanks! :D

[SoftTalker]

> For example the probability of getting [1,2,3,4,5,6] as the winning numbers in the lottery is the same as any random set of numbers.

Yes, but the comparison is not to "any random set of numbers" it's "all other random sets of numbers"

The candidate got 52.200000% of the vote instead of any other percentage, not another specific percentage.

[happyopossum]

> The candidate got 52.200000% of the vote instead of any other percentage, not another specific percentage

No, he got 51.1999971%. It’s right there in the second table of the article

[ithkuil]

Some other comment in this discussion claims that if you nudged the total count integer +-1 you'll never hit 52.20000000% exactly hinting at the possibility that they choose 52.2% exactly, then computed the total counts which would be a non-integer and then just rounded that.

Once re-computing the percentage from that number you end up with the slightly less round-looking 51.1999971%.

[neura]

I think you are correct, but that's missing the point of the article's content. I'm just a programmer, not a math expert, but I believe these statements are accurate.

1. It's very easy to arrive at the provided values, if you make up some percentages that only go to a single decimal value (1/10th). Though doing so would result in vote counts that are decimal, as well. Then if you just remove the decimal from those values, the given percentages don't change enough to be incorrect, but even when taken to 7 decimal places, the new values are pretty clearly due to the rounding (44.2%: 44.1999989%, 4.6%: 4.6000039%).

2. While yes, the chance of these vote counts coming up in this kind of pattern is similar to the example you provided, even if you were using 0-9 for your example of 6 values, the total combinations is about an order of magnitude less than the total vote count provided here.

3. The finer point made is that there's a very small chance for one of the vote counts to show up as a number that so nicely fits the single decimal percentage, but in this case, all 3 vote counts fit this pattern. The calculations are shown for just 2 of the candidates (so not including the "other") resulting only a 1 in 100 million chance.

[tumult]

That’s not what this article is about. Read the article.

[shadowgovt]

FWIW, statistics is hard.

Having read the article, I came away with similar questions and I appreciate the sibling comments clarifying.

[tumult]

Fair enough. Sorry.